The value of the integral sumlimits_{k = 1}^n | vedclass

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(C) Let $I_k = \int_0^1 {f(k - 1 + x)\,dx} $. Substitute $t = k - 1 + x$,then $dt = dx$. When $x = 0$,$t = k - 1$. When $x = 1$,$t = k$. So,$I_k = \in

Vedclass · Accepted Answer

$\int_0^n {f(x)\,dx} $

Vedclass · Answer

(C) Let $I_k = \int_0^1 {f(k - 1 + x)\,dx} $. Substitute $t = k - 1 + x$,then $dt = dx$. When $x = 0$,$t = k - 1$. When $x = 1$,$t = k$. So,$I_k = \int_{k - 1}^k {f(t)\,dt} = \int_{k - 1}^k {f(x)\,dx} $. Now,the given sum is $\sum\limits_{k = 1}^n {I_k} = \sum\limits_{k = 1}^n {\int_{k - 1}^k {f(x)\,dx} } $. Expanding the summation,we get: $\int_0^1 {f(x)\,dx} + \int_1^2 {f(x)\,dx} + \dots + \int_{n - 1}^n {f(x)\,dx} $. By the property of definite integrals,$\int_a^b f(x)dx + \int_b^c f(x)dx = \int_a^c f(x)dx$. Therefore,the sum equals $\int_0^n {f(x)\,dx} $.

The value of the integral $\sum\limits_{k = 1}^n {\int_0^1 {f(k - 1 + x)\,dx} } $ is

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